3\nthe diagram shows a container in the form of an inverted cone, with base radius 12 cm and\nheight 6 cm…

3\nthe diagram shows a container in the form of an inverted cone, with base radius 12 cm and\nheight 6 cm. the container is initially empty. at time ( t = 0 ), water is allowed to flow into the\ncontainer. after ( t ) seconds, the height of water in the container is ( (6 - x) ) cm and the volume of\nwater is ( v mathrm{cm}^{3} ), where ( x mathrm{cm} ) is the difference in height between the cone and the water level.\n(i) show that ( v=\frac{4 pi}{3}(6 - x)^{3} ).\ngiven that the rate of flow of the water is ( k(6 - x)^{2} mathrm{cm}^{3} / mathrm{s} ), where ( k ) is a constant,\n(ii) find an expression for the rate of change of the difference in height between the cone\nand the water level in terms of ( pi ) and ( k ).\nthe water from the tap was allowed to flow for ( 32 pi ) seconds when the height of the water\nreached 1 cm.\n(iii) find the value of ( k ).

3\nthe diagram shows a container in the form of an inverted cone, with base radius 12 cm and\nheight 6 cm. the container is initially empty. at time ( t = 0 ), water is allowed to flow into the\ncontainer. after ( t ) seconds, the height of water in the container is ( (6 - x) ) cm and the volume of\nwater is ( v mathrm{cm}^{3} ), where ( x mathrm{cm} ) is the difference in height between the cone and the water level.\n(i) show that ( v=\frac{4 pi}{3}(6 - x)^{3} ).\ngiven that the rate of flow of the water is ( k(6 - x)^{2} mathrm{cm}^{3} / mathrm{s} ), where ( k ) is a constant,\n(ii) find an expression for the rate of change of the difference in height between the cone\nand the water level in terms of ( pi ) and ( k ).\nthe water from the tap was allowed to flow for ( 32 pi ) seconds when the height of the water\nreached 1 cm.\n(iii) find the value of ( k ).

Answer

(i)

Explanation:

Step1: Use similar triangles

Let the radius of the water - surface be (r). By similar triangles, (\frac{r}{12}=\frac{6 - x}{6}), so (r = 2(6 - x)).

Step2: Use the volume formula for a cone

The volume of a cone (V=\frac{1}{3}\pi r^{2}h). Substitute (r = 2(6 - x)) and (h=(6 - x)) into the formula. [ \begin{align*} V&=\frac{1}{3}\pi(2(6 - x))^{2}(6 - x)\ &=\frac{1}{3}\pi\times4(6 - x)^{2}(6 - x)\ &=\frac{4\pi}{3}(6 - x)^{3} \end{align*} ]

(ii)

Explanation:

Step1: Differentiate (V) with respect to (x)

We know (V=\frac{4\pi}{3}(6 - x)^{3}). Using the chain - rule (\frac{dV}{dx}=4\pi(6 - x)^{2}).

Step2: Use the chain - rule (\frac{dx}{dt}=\frac{dx}{dV}\times\frac{dV}{dt})

Given (\frac{dV}{dt}=k(6 - x)^{2}). Since (\frac{dx}{dV}=\frac{1}{\frac{dV}{dx}}), then (\frac{dx}{dt}=\frac{k(6 - x)^{2}}{4\pi(6 - x)^{2}}).

Step3: Simplify the expression

Cancel out ((6 - x)^{2}) (assuming (x\neq6)), we get (\frac{dx}{dt}=\frac{k}{4\pi})

(iii)

Explanation:

Step1: Integrate (\frac{dx}{dt})

We know (\frac{dx}{dt}=\frac{k}{4\pi}), so (x=\frac{k}{4\pi}t + C). When (t = 0), (x=6) (since the container is initially empty, the height of water (h = 0), so (6-0=6)), then (C = 6). So (x=\frac{k}{4\pi}t+6). When (t = 32\pi), (h = 1) cm, then (x=6 - 1=5)

Step2: Substitute values into the equation

Substitute (t = 32\pi) and (x = 5) into (x=\frac{k}{4\pi}t+6) [ \begin{align*} 5&=\frac{k}{4\pi}\times32\pi+6\ 5&=8k + 6\ 8k&=- 1\ k&=-\frac{1}{8} \end{align*} ] Another way:

Step1: Use the volume formula

The volume of water when (h = 1) (i.e., (x = 5)) is (V=\frac{4\pi}{3}(6 - 5)^{3}=\frac{4\pi}{3}).

Step2: Use the formula (V=\int_{0}^{t}\frac{dV}{dt}dt)

Since (\frac{dV}{dt}=k(6 - x)^{2}) and (x = 6 - h), and (V=\int_{0}^{32\pi}k(6-(6 - h))^{2}dt) (but using (V=\int_{0}^{t}\frac{dV}{dt}dt) and (V=\frac{4\pi}{3}), (t = 32\pi), (\frac{dV}{dt}=k(6 - x)^{2}) and from (V=\frac{4\pi}{3}(6 - x)^{3}), when (x = 5), (V=\frac{4\pi}{3}) [ \begin{align*} \frac{4\pi}{3}&=\int_{0}^{32\pi}k(6 - x)^{2}dt\ \text{Since }V=\frac{4\pi}{3}(6 - x)^{3}\Rightarrow dV = 4\pi(6 - x)^{2}dx\ \text{and }\frac{dV}{dt}=k(6 - x)^{2}\Rightarrow dV=k(6 - x)^{2}dt\ \frac{4\pi}{3}&=\int_{0}^{32\pi}k(6 - x)^{2}dt\ \text{From (ii) }\frac{dx}{dt}=\frac{k}{4\pi}\Rightarrow(6 - x)^{2}dt=\frac{4\pi}{k}dx\ \frac{4\pi}{3}&=\int_{6}^{5}k\times\frac{4\pi}{k}dx\ \frac{4\pi}{3}&=4\pi\int_{6}^{5}dx\ \frac{4\pi}{3}&=4\pi(5 - 6)\ \text{This is wrong. Let's use the first method.}\ \end{align*} ] We know (x=\frac{k}{4\pi}t+6), when (t = 32\pi), (x = 5) [ \begin{align*} 5&=\frac{k}{4\pi}\times32\pi+6\ 5&=8k + 6\ 8k&=-1\ k&=-\frac{1}{8} \end{align*} ]

Answer:

(i) Shown as above. (ii) (\frac{dx}{dt}=\frac{k}{4\pi}) (iii) (k =-\frac{1}{8})